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I was tasked with finding $\min(\mathbb{Q},i+\sqrt[]{2})$

I found the polynomial in $\mathbb{Z}[X]$ to be $X^4-2X^2+9$

I know it must be minimal over $\mathbb{Q}$ since (used brute force) any arbitrary qubic, quadratic, or linear polynomial has coefficients in $\mathbb{C}-\mathbb{Q}$

My question is is there an easier method?

(Edit: question initially asked about irreducibility, but since minimal implies irreducibility, we make the change)

Bill Dubuque
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northcity4
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4 Answers4

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Eisenstein works very well!

Let $X=\frac{2}{x}+1.$

Thus, $$X^4-2X^2+9=\frac{8(x^4+2x^2+4x+2)}{x^4}.$$ Now, take $p=2$.

Because if $X^4-2X^2+9$ is reducible then $x^4+2x^2+4x+2$ is reducible, which is a contradiction by Eisenstein.

Michael Rozenberg
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  • This is a function in $\mathbb{Z}(X)$, not $\mathbb{Z}[X]$ I am not seeing why this even works as a trick for eisensteins criterion – northcity4 Feb 17 '18 at 03:37
  • Because if $X^4-2X^2+9$ is reducible then $x^4+2x^2+4x+2$ is reducible, which is a contradiction by Eisenstein. – Michael Rozenberg Feb 17 '18 at 03:40
  • But the choice of "a" as you used in the lemma I posted in the question needs to come from $\mathbb{Z}$ – northcity4 Feb 17 '18 at 03:46
  • My reasoning says that it's not necessary. See please better my reasoning. – Michael Rozenberg Feb 17 '18 at 03:54
  • I am sorry, but that is what the lemma says in this case. – northcity4 Feb 17 '18 at 04:13
  • If the lemma does not help, it does not say that we can't use Eisenstein here. – Michael Rozenberg Feb 17 '18 at 04:27
  • Eisenstein does say $x^4+2x^2+4x+2$ is irreducible over $\mathbb{Q}$, but I am not seeing how that relates to my polynomial proposed. If the relationship is through the substitution, I cannot use that – northcity4 Feb 17 '18 at 04:33
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    This is a good idea. It may (if the asker has learned about reciprocal polynomials) become more accessible to your readers if you split it into steps. First do the substitution $x=2t+1$ to show that irreductibility of this is equivalent to that of $$(2x+1)^4-2(2x+1)^2+9=16x^4+32x^3+16x^2+8=8(2x^4+4x^3+2x^2+1).$$ Then go to reciprocal to conclude that this (sans the eight) is irreducible because $$x^4\left(2(1/x)^4+4(1/x)^3+2(1/x)^2+1\right)=x^4+2x^2+4x+2 $$ is irreducible by Eisenstein. Not much to it here. It's just that the use of fractional linear transformations isn't always covered. – Jyrki Lahtonen Feb 17 '18 at 06:29
  • (cont'd) but may be the fact about using reciprocals was? Basically I'm using the fact that the transformations $x\mapsto 1/x$ and $x\mapsto x-a, a\in\Bbb{Q}$ generate the group of all the fractional linear transformations. Granted, this won't help the OP if they haven't done the exercise of showing that the reciprocal of an irreducible polynomial is also irreducible. – Jyrki Lahtonen Feb 17 '18 at 06:33
  • "Now, take $p=2$": what is "$p$"? – Kan't Dec 24 '24 at 11:09
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I will use a tower property in this case:

We know $K=\mathbb{Q}(i,\sqrt[]{2})$ is a field extension of $L=\mathbb{Q}(\sqrt[]{2})$ which is a field extension of $F=\mathbb{Q}$

In a previous result, we showed that $[K:L]=2$ and since $[L:Q]=2$ it follows $[K:Q]=4$

Therefore, the minimum polynomial must have degree 4

northcity4
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  • This is the easiest method in my opinion, but we need to show that $\mathbb{Q}(i+\sqrt{2}) = \mathbb{Q}(i,\sqrt{2})$. This however is easily seen if we notice that $(i+\sqrt{2})^2 \implies i\sqrt{2} \in \mathbb{Q}(i+\sqrt{2})$ and then calculate $i\sqrt{2}(i+\sqrt{2})$ . – The Tralfamadorian Jan 21 '23 at 09:23
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I fully endorse Michael's method. If you lack a key result and can't follow it, you can do one of the following.

  1. Show that $f(x)=x^4-2x^2+9$ is irreducible by observing that it has no rational roots, and eliminating a possible factorization into quadratics by concluding from $$f(x)=(x^2+ax+b)(x^2+cx+d)$$ that first (compare cubic terms) we must have $c=-a$, then (compare linear terms) also $b=d$. Therefore $b=d=\pm3$, and checking that neither sign works is easy.
  2. Examining the field generated by $\alpha=i+\sqrt2$. We have $$\frac1\alpha=\frac{\sqrt2-i}3,$$ so you can write both $\sqrt2$ and $i$ as $\Bbb{Q}$-linear combinations of $\alpha$ and $1/\alpha$. Hence $\alpha$ is a generator of the field $K=\Bbb{Q}(i,\sqrt2)$. As $[K:\Bbb{Q}]=4$ the minimal polynomial of $\alpha$ must have degree four.

The choice that you find acceptable depends heavily on what tools have been covered already in class. At this point in a typical algebra course the theory moves forward swiftly, so it is quite difficult for us to determine what really helps you. And in a few weeks time a tool not yet covered may be in frequent use.

Jyrki Lahtonen
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A variant of Michael Rozenberg’s method, broken down into several steps:

Step 1: $f(x)=x^4 -2x^2+9$ is irreducible if and only if $\,f(x+1)=x^4+4x^3+4x^2+8$ is irreducible. Call this $g(x)$.

Step 2: $g$ above is irreducible if and only if $8x^4+4x^2+4x+1$ is irreducible. Call this polynomial $h(x)$.

Step 3: $h$ above is irreducible if and only if $h(x/2)=\frac12x^4+x^2+2x+1$ is irreducible. And when you multiply this by $2$, you get an Eisenstein polynomial, irreducible.

So the original $f(x)$ is irreducible.

(Maybe I should say, for the benefit of the cognoscenti that if you know Newton Polygon, you can see irreducibility at Step 1, ’cause the Polygon passes through no integral points other than $(0,3)$ and $(4,0)$. This is exactly the phenomenon you use when you prove Eisenstein.)

Lubin
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